Difficult problem below, can anyone help?

Question

A study done for management shows that the revenue, R, in dollars, as a function of number, x, of units produced can be modeled by R(x) = 10x - 0.0005x^2 while the cost, C, in dollars, can be modeled by C(x) = 5000 + 2x. What will the maximum profit be, in dollars?

Answer 1

Profit:
P(x)=R(x)-C(x)
P(x)=10x-0.0005x^2-(5000+2x)
P(x)=10x-0.0005x^2-5000-2x
P(x)=8x-0.0005x^2-5000

Maximum Profit:
P'(x)=0
8-0.001x=0
8=0.001x
x=$8,000

I hope this helps!

Answer 2

The profit, in terms of the number of units, is

P(x) = R(x) - C(x) = 10x - 0.0005x^2 - 5000 - 2x
= 8x - 0.0005x^2 - 5000.

To find the maximum profit, find the value of x for which P'(x) = 0. You have

P'(x) = 8 - 0.001x, so P'(x) = 0 when x = 8/0.001 = 8000.

This is a maximum, because:
1) P''(8000) = -0.001 < 0, so P has a local maximum at x=8000.
2) P(0) = -5000 and P(x) -> -infinity as x -> infinity, so this local maximum must be the absolute maximum for x in the interval [0,infinity).

The maximum profit will be P(8000).

Answer 3

P(x)= R(x)-C(x)

=10x-0.0005x^2-5000 -2x

= -0.0005x^2+8x-5000

the max profit will occur when dP(x)/dx =0,
since the second differential of P(x)
= -0.001 which is -ve

dP(x)/dx = -0.001x+8= 0,

>>> x=8000 at max profit

P(8000)= -0.0005(8000)^2
+8*8000-5000

=27,000

therefore, maximum profit =$27,000

Answer 4

ask your parents' help with your homework...stop cheating, it will get you no where.

And anyone who gives them the answer is nothing more than a cheat themselves.

Answer 5

maximum profit will be [R(y)*x - C(y)/2x+y] $

Answer 6

do your home work

Answer 7

i would help u but i live in sullivanIL